balok sederhana

Konstruksi Balok Sederhana

Contoh soal dan penyelesaian MK Statika pelengkung tiga sendi

                                                q= 1 t/m

                                                            S                                                 

                                                                                                                                c=4m

  A                                                                                           B            

 

                                a=4m                                     b=4m

                                                L= 8m

Ditanyakan ;       a. reaksi perletakan

                                b. bidang M,D,N

Penyelesaian:

                Analisis

q= 1 t/m

                                                            S                                                 

                                                                                                                                c=4m

   RAH                                                                                                    RBH

  A                                                                                           B

                   RAV                                                                    RBV       

 

                                a=4m                                     b=4m

                                                L= 8m

a.       Reaksi perletakan

Yang perlu selalu diingat dalam penyelasaian kasus-kasus pada statika tertentu adalah :

∑M=0

∑V=0

∑H=0

Dalam kasus ini yang diselesaikan adalah :

∑MA=0

∑MB=0

∑MS=0 (bagian A-S atau B-S, dan salah satunya digunakan untuk control atau mengecek apakah sudah ok apa belum)

Q=q.L = 1 t/m . 8m = 8t

∑MA=0

       0 = -RBV.L + Q.a

       0 = - RBV.8 + 8.4

 RBV = 32/8 = 4t

∑MB=0

       0 = -RAV.L + Q.b

       0 = - RAV.8 + 8.4

 RAV = 32/8 = 4t

∑MS=0                  (bentang A-S)

       0 = - q.a. ½ a + RAV.a – RAH.c

       0 = - 1.4. ½ 4 + 4.4 – RAH.4

       0 = - 8 + 16 – RAH.4

 RAH = 8/4 = 2t (         )

∑H=0    

       0 =  RAH – RBH

       0 = 2  – RBH

 RBH =  2t (         )

Control

∑MS=0                  (bentang B-S)

       0 = - q.b. ½ b + RBV.b – RBH.c

       0 = - 1.4. ½ 4 + 4.4 – RBH.4

       0 = - 8 + 16 – 2.4

       0 = 0  ( ok )

                q= 1 t/m

                                                            S                                                 

                                                                                                                                c=4m

   RAH = 2t                                                                                            RBH=2t

  A                                                                                           B

                   RAV= 4t                                                                             RBV= 4t               

 

                                a=4m                                     b=4m

                                                L= 8m

b.      Bidang M,D,N

Untuk menyelesaikan bidang M,D,N selalu mengadakan Potongan…dan peninjauan

Adakan potongan sejauh x dari A (bentang A-S) dan tinjau kiri

Tg α= 4/4 = 1, α= 45⁰

                qx

                                                         y= x tg α

       RAH= 2t         45⁰                                      

                                X

                RAV= 4t

                qx= q.x = x

                Mx = +RAV.x – qx. ½ x – RAH.y

                       = + 4x – ½ x²  – 2y      0≤ x ≤ 4

                X=0        Mx = 0

                X=4        Mx = 0

                Mx max atau titik puncak momen

                dMx/dx = 0 = 4 – x – 2

                 puncak saat x = 2 dari A

                Mx max = 4.2 – ½ 2² – 2.2 =  2 tm

               

Untuk bidang D

                Tg α= 4/4 = 1, α= 45⁰

                qx

            RAH sinα                                          

 RAH= 2t               45⁰                                      

                                X

      RAV= 4t          RAV cos α

                Dx= RAVcosα – RAHsinα – qxcos²α

                     = 4. ½ √2 – 2. ½ √2 – x. ½

                     = 2√2 - √2 – ½ x            0≤x≤4

                X=0        Dx= √2= 1,41

                X=4        Dx = √2 – 2= - 0,59

                Untuk bidang N

                                qx

                                                        

       RAH= 2t         45⁰                                      

RAHcosα

+ RAVsinα                           X

                RAV= 4t

                Nx = RAHcosα+ RAVsinα – qxsin²α

                      = 2. ½ √2 + 4. ½ √2 – ½ x

                     = √2 + 2√2 – ½ x

                     = 3√2 – ½ x                     0 ≤x ≤4

                x=0         Nx = 3√2=4,23

                x=4         Nx = 2,23

               

                karena kedua bentang sama (A-S dan B-S).. maka tidak perlu susah-susah menghitung lagi karena hasilnya akan sama. Hanya saja tandanya akan berbeda dalam proses penyelesaian.

                                                                                                                q= 1 t/m

                                                                                                                S

                                                                                                                                                                                                                                                                C=4m

RAH = 2t                                                                                                                                                                                                               RBH=2t

                                                A                                                                                                                                     B

                                                RAV= 4t                                                                                                                                RBV=4t

                                                                a=4m                                                                     b=4m

                                                                                +                                                             +                                                             Bidang Momen (M)

-                        -

       +                                                                                                                      +                             Bidang Gaya Lintang (D)

                                                                                +                                                                             +                                             Bidang Gaya Normal (N)